一个简单的查询 [英] a simple query
问题描述
printf（"％d"，（float）3/2）;
输出为零...为什么不是1？即使你把它改成
printf（"％d"，（float）6/2）;仍然输出为零
printf("%d",(float)3/2);
the output comes as zero...why not 1?even when u change it to
printf("%d",(float)6/2); still the output is zero
推荐答案
newbai写道：
newbai wrote:
printf （"％d"，（float）3/2）;
输出为零...为什么不是1？即使你把它更改为
printf （QUOT;％d"，（浮点）6/2）;仍然输出为零
printf("%d",(float)3/2);
the output comes as zero...why not 1?even when u change it to
printf("%d",(float)6/2); still the output is zero
Who''su？
％d告诉printf你传递一个整数值，但是你是通过一个浮动来
，那是'未定义的行为。
-
Ian Collins。
Who''s u?
%d is telling printf you are passing an integer value, but you are
passing a float, that''s undefined behaviour.
--
Ian Collins.
newbai写道：
newbai wrote:
printf（"％d"，（float）3/2）;
输出为零...为什么不是1？即使你把它更改为
printf（"％d"，（float）6/2）;仍然输出为零
printf("%d",(float)3/2);
the output comes as zero...why not 1?even when u change it to
printf("%d",(float)6/2); still the output is zero
％d格式说明符用于打印int参数。给printf
错误的类型会导致不明确的行为，即任何事情都可以发生
。
做：
printf（"％d \ n"，3/2）;和printf（％d \ n，6/2）;
或
printf（"％f \ n，（浮动）3/2）;和printf（"％f\ n"，（float）6/2）;
The %d format specifier is for printing int arguments. Giving printf
the wrong type will cause undefined behaviour, i.e. anything can
happen.
Do:
printf("%d\n", 3/2); and printf("%d\n", 6/2);
or
printf("%f\n", (float)3/2); and printf("%f\n", (float)6/2);
newbai写道：
newbai wrote:
printf（"％d"，（float）3/2）;
输出为零...为什么不是1？即使你把它改为
printf（"％d"，（float）6/2）;仍然输出为零
printf("%d",(float)3/2);
the output comes as zero...why not 1?even when u change it to
printf("%d",(float)6/2); still the output is zero
％d想要一个int，表达式是一个浮点数，所有投注都关闭。
明天BOOM。明天总会有一个BOOM。
-
Chris" Ivanova" Dollin
你服务的对象是谁，你信任谁？ /十字军/
%d wants an int, the expression is a float, all bets are off.
BOOM tomorrow. There''s always a BOOM tomorrow.
--
Chris "Ivanova" Dollin
"Who do you serve, and who do you trust?" /Crusade/
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